The following initial rate data were found for the reaction 2MnO^-_4 + 5H_2C_2O_
ID: 998171 • Letter: T
Question
The following initial rate data were found for the reaction 2MnO^-_4 + 5H_2C_2O_4 + 6H^+ rightarrow 2Mn^2+ + 10CO_2 + 8H_2O which of the following is the correct rate law? Rate = k[MnO^-_4]^2[H_2C_2O_4]^5[H^+]^6 Rate = k[MnO^-_4]^2[H_2C_2O_4][H^+] Rate = k[MnO^-_4][H_2C_2O_4][H^+] Rate = k[MnO^-_4]^2[H_2C_2O_4] Rate = k[MnO^-_4]^2[H_2C_2O_4]^2 What is the expected boiling point of a solution prepared by dissolving 6.37 g of sodium iodide (Nal) in 48.6 g of water (H_2O)? For water, T_b = 100.00 degree C and K_b = 0.512 degree C m^-1. A 0.20 M solution of Mg(So_4) has an observed osmotic pressure of 7.6 atm at 25 degree C. Determine the observed van't Hoff(i) factor for this experiment. 19 0.31 1.6 1.8 2.0Explanation / Answer
21) when we check the data of initial rate, it is found that ,
on doubling the initial concentration of MnO4-, keeping the conc. of other reactants constant, the rate of the reaction increased four times. So it is a second order reaction with respect to [MnO4-]
On doubling the initial concentration of H2C2O4, keeping the conc. of other reactants constant, the rate of the reaction increased two times. So it is a first order reaction with respect to [H2C2O4]
On doubling the initial concentration of H+, keeping the conc. of other reactants constant, the rate of the reaction did not change. So it is a zero order reaction with respect to [H+]
Then rate law can be expressed as rate= k [MnO4-]2[H2C2O4]
22) Using the eqn, T = i Kb m
where i is van 't Hoff factor , which is 2 for NaI
m is the molality of the solution = 6.37g/149.89 g/mol/.0486 Kg = 0.874 m
Kb= 0.512 oC/m
T= 2 x 0.512 oC/m x 0.874 m = 0.8949 oC
Therefore, boiling point of the solution is Tb+0.8949 oC= 100 +0.8949 oC = 100.8949 oC
23)
Using the osmosis equation:
= i M R T where,
osmotic pressure () = 7.6 atm, molarity (M) = 0.2 M , gas constant (R) = 0.08206 L atm mol¯1 K¯1 andtemperature (T) = 25 °C = 298 K
Then i = /MRT = 7.6 atm /( 0.2 M x 0.08206 L atm mol¯1 K¯1 x 298 K)
= 1.6
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